Reflection and refraction: follow the ray at a boundary

MCAT content library / Chemistry & physics / Reflection and refraction: follow the ray at a boundary

A ray’s angle, its incident medium, and the index ratio determine whether it bends toward the normal, bends away, or reflects internally.

About 21 minutes, including the worked examples and checks.

Useful foundations

• Waves: separate amplitude, frequency, and speed
• Electromagnetic light: wavelength, photons, and polarization

On this page

• Begin with the boundary and its normal
• Use refractive index to connect speed and direction
• Worked application: entering a transparent block
• Identify when Snell’s law reaches its limit
• Apply the condition to an optical fiber
• Dispersion means the index depends on frequency
• Check your reasoning
• Continue learning
• Sources

Begin with the boundary and its normal

A light ray reaching a boundary can produce both reflected and transmitted light. The reflected ray stays in the incident medium. A transmitted ray enters the second medium, where its direction can change. These geometrical outcomes depend on the surface orientation and the refractive indices at the light’s frequency. AAMC category 4D includes reflection, refraction, dispersion, and total internal reflection.¹

The normal is a line perpendicular to the surface at the point where the ray strikes it. Measure the incident, reflected, and refracted angles from this normal. A ray at 20° to the surface is at 70° to the normal. Confusing those complementary angles reverses many predictions before any equation is used.

For a smooth planar boundary, the angle of reflection equals the angle of incidence. The incident ray, normal, and reflected ray lie in one plane. Reflection does not require the second medium to have a lower refractive index, and ordinary reflection can coexist with transmission. Total internal reflection is a more restrictive condition.²

Use refractive index to connect speed and direction

The refractive index is n = c/v, where c is vacuum light speed and v is the phase speed in the material at the relevant frequency. For the ordinary transparent materials considered here, a larger n means a slower speed. At a stationary interface, the transmitted light retains the incident frequency; its wavelength changes with speed. Thus λ_medium = λ_vacuum/n.³

Snell’s law is n₁sinθ₁ = n₂sinθ₂. Subscripts identify the incident and transmitted media, not whichever side happens to be drawn above the other. Entering a higher-index material bends an oblique transmitted ray toward the normal. Entering a lower-index material bends it away, when a propagating transmitted ray is possible. At normal incidence, both angles are zero and no angular bending occurs even though speed changes.

Worked application: entering a transparent block

The indices, angles, and ray diagrams in this example are original illustrative inputs. Take air as n = 1.00 and the block as n = 1.50 at one frequency. A ray arrives at 30.0° to the normal. Then sinθ₂ = (1.00/1.50)sin 30.0° = 1/3, giving θ₂ ≈ 19.5°. The reflected ray is at 30.0°. The transmitted light’s speed is c/1.50, but neither ray changes frequency merely because the boundary is stationary.

Case	Incident → second index	Incident angle	Result
Entry	1.00 → 1.50	30.0°	Refraction at 19.5°; reflection at 30.0°
Exit below critical	1.50 → 1.00	30.0°	Refraction at 48.6°
Exit above critical	1.50 → 1.00	50.0°	Total internal reflection at 50.0°

On the lower-angle exit, sinθ₂ = 1.50sin30.0° = 0.750, so θ₂ ≈ 48.6°. The path bends away from the normal. Reversing the complete entry path is also possible: an internal ray at 19.5° exits into air at 30.0°. Keeping the same numerical incident angle after swapping materials is a different experiment, not a reversed path.

Identify when Snell’s law reaches its limit

When n₁ > n₂, increasing θ₁ eventually makes the refracted angle reach 90°. The corresponding critical angle satisfies sinθ_c = n₂/n₁. At θ_c, the limiting refracted direction runs along the boundary. For θ₁ > θ_c, Snell’s law would demand a sine greater than one, so there is no propagating refracted ray into the second medium: total internal reflection occurs in this ideal model.⁴

For the block–air boundary, θ_c = sin⁻¹(1.00/1.50) ≈ 41.8°. The 50.0° ray therefore reflects internally, whereas the 30.0° ray can transmit. A large incident angle alone is insufficient. A ray entering from air into this block cannot undergo total internal reflection at that entry boundary, because it starts in the lower-index medium.

A wave description still includes an evanescent field extending a short distance beyond the interface. The ray statement means no ordinary propagating transmitted ray in the second medium, not that the electromagnetic field is zero at every point beyond the boundary. Calculating that field is unnecessary for the geometrical prediction here.⁶

Apply the condition to an optical fiber

Consider a straight illustrative fiber with core index 1.50 and surrounding cladding index 1.20. At their interface, sinθ_c = 1.20/1.50 = 0.800, so θ_c ≈ 53.1°. An internal ray striking at 60.0° to the normal satisfies the condition for total internal reflection. A 45.0° ray does not. For this straight boundary, a 30.0° angle to the fiber’s axis corresponds to 60.0° to its normal; the two angle conventions must not be interchanged.

This calculation concerns an already established internal ray. It does not by itself give the external launch-angle range, include losses from bending, or guarantee perfect transmission through a real fiber. Those require additional geometry or material information.

Dispersion means the index depends on frequency

Suppose a second block has n = 1.60 for a blue component and n = 1.50 for a red component. Each enters from air at 30.0°. The predicted refracted angles are sin⁻¹(0.500/1.60) ≈ 18.2° for blue and 19.5° for red. The higher-index component bends closer to the normal. This separation follows from the supplied index values, rather than from a rule that one color must always bend more in every possible material.⁵

The frequencies remain those supplied by the source. Dispersion changes the components’ speeds and wavelengths within the material and can separate their directions. It does not create new colors by converting one frequency into another. For any boundary question, first mark the normal, then identify the incident medium, and finally ask whether the transmitted-angle calculation gives a physically possible sine.

Check your reasoning

Choose an answer before opening its feedback. Each option explains the reasoning behind that choice.

Continue learning

Related lessons

• Waves: separate amplitude, frequency, and speed
• Electromagnetic light: wavelength, photons, and polarization

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Sources

1. AAMC: C/P content category 4D

2. OpenStax University Physics 3: The Law of Reflection

3. OpenStax University Physics 3: Refraction

4. OpenStax University Physics 3: Total Internal Reflection

5. OpenStax University Physics 3: Dispersion

6. MIT: Electromagnetic Field Theory, Chapter 7